Inverse of the matrix product $\boldsymbol{A} \cdot\boldsymbol{S} \cdot \boldsymbol{A}^{T}$

Question

If I have an $n\times n$ symmetric matrix $\boldsymbol{S}$ and a $m\times n$ matrix $\boldsymbol{A}$ is there any relation between $(\boldsymbol{A} \cdot\boldsymbol{S} \cdot \boldsymbol{A}^{T})^{-1}$ and $\boldsymbol{S}^{-1}$?

Answer 1

We begin by noting that $m \leq n$, and $A$ has rank $m$.

Because $A$ has full rank, there exists an invertible $n \times n$ matrix $B$ such that $$ AB^{-1} = \pmatrix{I_m&0} : = J $$ (we can find this matrix by row-reducing $A^T$). We then have $$ (ASA^T)^{-1} = (JB^SB^{T}J^T)^{-1} = ((B SB^{T})[\{1,\dots,m\}])^{-1} $$ Where $M[S]:= [m_{ij}]_{i,j \in S}$. That is, we are finding the inverse of a submatrix of $BSB^T$.

Because there is no nice general relationship between the inverse of a submatrix and the inverse of the entire matrix, there is no nice relationship between $((B SB^{T})[\{1,\dots,m\}])^{-1}$ and $(BSB^T)^{-1} = B^{-T}S^{-1}B^{-1}$, so there is no nice relationship between $(ASA^T)^{-1}$ and $S^{-1}$.